OFFSET
1,4
COMMENTS
If p is prime, then a(p) = 1. The only pair of divisors (d1,d2) of p such that d1 < d2 is (1,p), and since p^2 + 1*(p - p)^2 = p^2, (1,p) is the only solution.
FORMULA
a(n) = Sum_{d1|n, d2|n, d1<d2} [d2^2 + d1*(n - d2)^2 = n^2], where [ ] is the Iverson bracket.
EXAMPLE
a(12) = 7; The 7 pairs are (1,12), (2,4), (2,12), (3,6), (3,12), (4,12), (6,12).
MATHEMATICA
Table[Sum[Sum[KroneckerDelta[k^2 + i*(n - k)^2, n^2] (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k - 1}], {k, n}], {n, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Wesley Ivan Hurt, May 19 2021
STATUS
approved